Problem: $h(n) = n-5$ $f(t) = 4t^{2}-4+5(g(t))$ $g(t) = -t+6-4(h(t))$ $ h(f(7)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(7)$ . Then we'll know what to plug into the outer function. $f(7) = 4(7^{2})-4+5(g(7))$ To solve for the value of $f$ , we need to solve for the value of $g(7)$ $g(7) = -7+6-4(h(7))$ To solve for the value of $g$ , we need to solve for the value of $h(7)$ $h(7) = 7-5$ $h(7) = 2$ That means $g(7) = -7+6+(-4)(2)$ $g(7) = -9$ That means $f(7) = 4(7^{2})-4+(5)(-9)$ $f(7) = 147$ Now we know that $f(7) = 147$ . Let's solve for $h(f(7))$ , which is $h(147)$ $h(147) = 147-5$ $h(147) = 142$